# Show that two series are equal

In my harmonic analysis class I have to prove that for all $a>0$ the following equality holds: $$\sum_{n\in \mathbb{Z}}e^{-n^2\pi a}=\frac{1}{\sqrt{a}}\sum_{k\in \mathbb{Z}}e^{-k^2 \pi / a}.$$

I'd appreciate any help on how to approach this problem and where to start.

Thank you!

Edited

I tried to apply Poisson's summation formula. I guessed that the function is $$f(x)=\exp(-\frac{x^2a}{4\pi}).$$ On the right I get $\sum_{n\in \mathbb{Z}}e^{-n^2\pi a}$, as required. But for the left side I have to calculate the integral $$\int_{-\infty}^{\infty}f(x)\exp(-i\omega x) dx ,$$ and I can't find the right substitution to solve it.

Any help would be appreciated!

• Poisson summation formula – Daniel Fischer Jan 27 '14 at 19:40
• – Mhenni Benghorbal Jan 27 '14 at 19:52
• @DanielFischer: thank you very much! I don't know how I missed that! – Ludolila Jan 27 '14 at 20:17
• @MhenniBenghorbal:thanks! It helped a lot! – Ludolila Jan 27 '14 at 20:18
• @Ludolila: You are very welcome. – Mhenni Benghorbal Jan 27 '14 at 20:19

The main idea is to use Poisson summation formula: $$\sum_{k\in \mathbb Z} f(2\pi k) = \frac{1}{2\pi} \sum_{n\in \mathbb Z} \hat{f}(n).$$ We start with setting $f(x)=e^{-\frac{x^2}{4\pi a}}$, and thus $f(2\pi k)= e^{-\frac{k^2\pi}{a}}.$ Now we have to find $\hat{f}(\omega)$. At first I thought to solve the above-mentioned integral, but its not that easy. So I took a different approach (thought I couldn't remember where I've seen it before): we can differentiate $f(x)$ to get a differential equation $$f'(x)+\frac{x}{2\pi a} f(x)=0.$$ Applying Fourier transform to this equation we get $$\omega \hat{f}(\omega)+\frac{1}{2\pi a} (\hat{f})'(\omega).$$ Solving this diff. equation we get $\hat{f}(\omega)=2\pi\sqrt{a}e^{-\pi a \omega^2}$ and that proves the claim.